I'm reading part 3. of below theorem:
Summary: Let $\left(\Omega_{1}, \mathcal{F}_{1}\right)$ and $\left(\Omega_{2}, \mathcal{F}_{2}\right)$ be two measurable spaces, and $\mathcal{F}_{1} \otimes \mathcal{F}$ their product $\sigma$-algebra. If $f: \Omega_{1} \times \Omega_{2} \rightarrow \mathbb{R}$ is $\mathcal{F}_{1} \otimes \mathcal{F}_{2}$-measurable, then $f\left(\omega_{1}, \cdot\right): \Omega_{2} \rightarrow \mathbb{R}$ is $\mathcal{F}_{2}$-measurable for all $\omega_{1}$, and $f\left(\cdot, \omega_{2}\right): \Omega_{1} \rightarrow \mathbb{R}$ is $\mathcal{F}_{1}$-measurable for all $\omega_{2}$.
My understanding: Let $\rho_{1}: \Omega_{1} \times \Omega_{2} \rightarrow \Omega_{1}$ and $\rho_{2}: \Omega_{1} \times \Omega_{2} \rightarrow \Omega_{2}$ be the projection maps. Then $\rho_1, \rho_2$ are measurable. Let $f_1 (\cdot) := f\left(\omega_{1}, \cdot\right)$. For a Borel set $B \subseteq \mathbb R$, $$f_1^{-1} (B) = \rho_2 (f^{-1} (B) \bigcap ( \{\omega_1\} \times \Omega_2 )).$$
I guess $\{\omega_1\} \times \Omega_2$ and the image of a measurable set through $\rho_1$ are not neccessarily measurable. So I think the statement given by the author may not be correct. Could you clarify my confusion?
Best Answer
First notice that even though a set $A$ may not be measurable, you can considere $\rho_2(A)$. Second see that by definition $f^{-1}(B)$ is measurable and that $\{ \omega_1 \} \times \Omega_2$ is measurable if and only if $\{ \omega_1 \} \in \mathcal F _1$. So what you have written is correct and does not contradict the theory, you just don't see why it brings the measurability of $f_1$.
For $B \in \mathcal B(\mathbb R)$, $$ f_1^{-1}(B) = \{ \omega_2 \in \Omega_2 : f(\omega_1,\omega_2) \in B \} = \{ \omega_2 \in \Omega_2 : (\omega_1,\omega_2) \in f^{-1}(B) \} $$ which is a section. For $C \in \mathcal F _1 \otimes \mathcal F _2$ and $\omega_1 \in \Omega_1$ the section of $C$ with $\omega_1$ is defined as $$ C_{\omega_1} = \{ \omega_2 \in \Omega_2 : (\omega_1, \omega_2) \in C \}. $$ You can similarly define section with $\omega_2 \in \Omega_2$. All sections are measurable, this is a property of the product sigma algebra.
Proof : Let $\omega_1 \in \Omega_1$ be fixed, the case of a section in the other coordinate is similar. Denote $$ \mathcal C = \{ C \in \mathcal F_1 \otimes \mathcal F_2 : C_{\omega_1} \in \mathcal F_ 2\}, $$ then you only need to show that $\mathcal C$ is a sigma algebra containing the elementary rectangles to have that $\mathcal C = \mathcal F_1 \otimes \mathcal F_2$. The second point is obvious because when $A \times B$ is an elementary rectangle $$ (A \times B)_{\omega_1} = \left\lbrace \begin{array}{cc} \emptyset & if~\omega_1 \notin A \\ B & if~\omega_1 \in A\\ \end{array} \right. \in \mathcal F_2. $$ For the first point, see that $$ (C_{\omega_1})^c = (C^c)_{\omega_1} $$ and $$ (\cup_nC_n)_{\omega_1} = \cup_n (C_n)_{\omega_1}.\square $$ Although I really don't like this way of seeing that $f(\omega_1, \cdot)$ is measurable, I prefere to write $$ f(\omega_1, \cdot) = f \circ g,~~g(\omega_2) = (\omega_1,\omega_2) $$ and see that all coordinates of $g$ are measurable so $f(\omega_1, \cdot)$ is measurable as a composition of measurable maps.